// 几何综合 - 初中数学综合题型（50题）
export const 几何综合_QUESTIONS = [
  // 难度1-2：基础综合题目 (1-20)
  {
    stem: '在△ABC中，∠A = 60°，∠B = 80°，求∠C的度数。',
    difficulty: 1,
    answer: [40],
    hint1: '三角形内角和为180°',
    hint2: '∠C = 180° - 60° - 80°',
    solution: '【解题】\\n∠A + ∠B + ∠C = 180°\\n60° + 80° + ∠C = 180°\\n∠C = 180° - 140° = 40°\\n\\n答：∠C = 40°。'
  },

  {
    stem: '等腰三角形的顶角为40°，求底角的度数。',
    difficulty: 1,
    answer: [70],
    hint1: '两底角相等',
    hint2: '(180° - 40°) ÷ 2',
    solution: '【解题】\\n设底角为x°\\n2x + 40° = 180°\\n2x = 140°\\nx = 70°\\n\\n答：底角为70°。'
  },

  {
    stem: '平行四边形ABCD中，∠A = 70°，求∠B、∠C、∠D的度数。',
    difficulty: 2,
    answer: [110, 70, 110],
    hint1: '邻角互补，对角相等',
    hint2: '∠B = 180° - 70°',
    solution: '【解题】\\n∵ ABCD是平行四边形\\n∴ ∠A + ∠B = 180°（邻角互补）\\n∠B = 180° - 70° = 110°\\n\\n∠C = ∠A = 70°（对角相等）\\n∠D = ∠B = 110°\\n\\n答：∠B = 110°，∠C = 70°，∠D = 110°。'
  },

  {
    stem: '矩形的对角线长为10cm，一边长为6cm，求另一边长。',
    difficulty: 2,
    answer: [8],
    hint1: '矩形对角线相等',
    hint2: '用勾股定理',
    solution: '【解题】\\n设另一边长为x cm\\n\\n由勾股定理：\\n6² + x² = 10²\\n36 + x² = 100\\nx² = 64\\nx = 8\\n\\n答：另一边长为8cm。'
  },

  {
    stem: '圆的半径为5cm，求圆的周长和面积。',
    difficulty: 1,
    answer: [31.4, 78.5],
    hint1: 'C = 2πr，S = πr²',
    hint2: 'π ≈ 3.14',
    solution: '【解题】\\nr = 5cm\\n\\n周长C = 2πr = 2×3.14×5 = 31.4cm\\n面积S = πr² = 3.14×5² = 78.5cm²\\n\\n答：周长31.4cm，面积78.5cm²。'
  },

  {
    stem: '正方形的对角线长为4√2cm，求正方形的边长和面积。',
    difficulty: 2,
    answer: [4, 16],
    hint1: '正方形对角线 = 边长×√2',
    hint2: '边长 = 4√2 ÷ √2',
    solution: '【解题】\\n设边长为a cm\\n\\n对角线 = a√2\\na√2 = 4√2\\na = 4\\n\\n面积S = a² = 16cm²\\n\\n答：边长4cm，面积16cm²。'
  },

  {
    stem: '等边三角形的边长为6cm，求其高和面积。',
    difficulty: 2,
    answer: [5.2, 15.6],
    hint1: '高 = (√3/2)×边长',
    hint2: '面积 = (√3/4)×边长²',
    solution: '【解题】\\n边长a = 6cm\\n\\n高h = (√3/2)×6 = 3√3 ≈ 5.2cm\\n\\n面积S = (√3/4)×6² = 9√3 ≈ 15.6cm²\\n\\n答：高约5.2cm，面积约15.6cm²。'
  },

  {
    stem: '梯形的上底为3cm，下底为7cm，高为4cm，求面积。',
    difficulty: 1,
    answer: [20],
    hint1: 'S = (上底+下底)×高÷2',
    hint2: '(3+7)×4÷2',
    solution: '【解题】\\nS = (a+b)h/2\\n= (3+7)×4÷2\\n= 10×4÷2\\n= 20cm²\\n\\n答：面积为20cm²。'
  },

  {
    stem: '菱形的对角线长分别为6cm和8cm，求菱形的面积和边长。',
    difficulty: 2,
    answer: [24, 5],
    hint1: '面积 = 对角线乘积÷2',
    hint2: '边长用勾股定理',
    solution: '【解题】\\n对角线d₁ = 6cm，d₂ = 8cm\\n\\n面积S = d₁×d₂/2 = 6×8/2 = 24cm²\\n\\n边长a = √[(d₁/2)² + (d₂/2)²]\\n= √[3² + 4²]\\n= √25 = 5cm\\n\\n答：面积24cm²，边长5cm。'
  },

  {
    stem: '圆锥的底面半径为3cm，高为4cm，求母线长。',
    difficulty: 2,
    answer: [5],
    hint1: '母线、半径、高构成直角三角形',
    hint2: 'l² = r² + h²',
    solution: '【解题】\\nr = 3cm，h = 4cm\\n\\n母线l = √(r² + h²)\\n= √(3² + 4²)\\n= √25 = 5cm\\n\\n答：母线长5cm。'
  },

  {
    stem: '在Rt△ABC中，∠C = 90°，AC = 3，BC = 4，求斜边AB和斜边上的高。',
    difficulty: 2,
    answer: [5, 2.4],
    hint1: 'AB² = AC² + BC²',
    hint2: '面积法求高',
    solution: '【解题】\\nAB = √(AC² + BC²)\\n= √(3² + 4²)\\n= √25 = 5\\n\\n设斜边上的高为h\\nS = (1/2)×AC×BC = (1/2)×AB×h\\n(1/2)×3×4 = (1/2)×5×h\\n6 = 2.5h\\nh = 2.4\\n\\n答：AB = 5，高为2.4。'
  },

  {
    stem: '正六边形的边长为2cm，求其周长。',
    difficulty: 1,
    answer: [12],
    hint1: '正六边形有6条边',
    hint2: '周长 = 6×边长',
    solution: '【解题】\\n正六边形有6条边\\n边长 = 2cm\\n\\n周长 = 6×2 = 12cm\\n\\n答：周长为12cm。'
  },

  {
    stem: '两圆半径分别为3cm和5cm，圆心距为8cm，判断两圆的位置关系。',
    difficulty: 2,
    answer: ['外离'],
    hint1: '比较圆心距与半径和、差的关系',
    hint2: 'd = 8，r₁+r₂ = 8',
    solution: '【解题】\\nr₁ = 3cm，r₂ = 5cm，d = 8cm\\n\\nr₁ + r₂ = 3 + 5 = 8cm\\n\\n∵ d = r₁ + r₂\\n∴ 两圆外切\\n\\n答：两圆外切。'
  },

  {
    stem: '扇形的半径为6cm，圆心角为60°，求扇形的弧长和面积。',
    difficulty: 2,
    answer: [6.28, 18.84],
    hint1: 'l = nπr/180',
    hint2: 'S = nπr²/360',
    solution: '【解题】\\nr = 6cm，n = 60°\\n\\n弧长l = nπr/180\\n= 60×3.14×6/180\\n= 6.28cm\\n\\n面积S = nπr²/360\\n= 60×3.14×36/360\\n= 18.84cm²\\n\\n答：弧长6.28cm，面积18.84cm²。'
  },

  {
    stem: '三角形的三边长分别为3、4、5，判断三角形的形状。',
    difficulty: 1,
    answer: ['直角三角形'],
    hint1: '检验是否满足勾股定理',
    hint2: '3² + 4² = 5²',
    solution: '【解题】\\n3² + 4² = 9 + 16 = 25 = 5²\\n\\n∵ 满足勾股定理\\n∴ 这是直角三角形\\n\\n答：直角三角形。'
  },

  {
    stem: '等腰梯形的上底为4cm，下底为10cm，腰长为5cm，求高。',
    difficulty: 3,
    answer: [4],
    hint1: '作高，形成直角三角形',
    hint2: '底边差的一半为直角三角形的一条直角边',
    solution: '【解题】\\n作高，底边差 = 10 - 4 = 6cm\\n直角三角形的一条直角边 = 6/2 = 3cm\\n\\n由勾股定理：\\nh² + 3² = 5²\\nh² = 25 - 9 = 16\\nh = 4cm\\n\\n答：高为4cm。'
  },

  {
    stem: '圆内接正三角形的边长为6cm，求圆的半径。',
    difficulty: 3,
    answer: [3.46],
    hint1: 'R = a/√3',
    hint2: 'R = 6/√3 = 2√3',
    solution: '【解题】\\n正三角形边长a = 6cm\\n\\n外接圆半径R = a/√3\\n= 6/√3\\n= 2√3\\n≈ 3.46cm\\n\\n答：半径约3.46cm。'
  },

  {
    stem: '长方体的长、宽、高分别为3cm、4cm、5cm，求其体对角线长。',
    difficulty: 2,
    answer: [7.07],
    hint1: '体对角线² = 长² + 宽² + 高²',
    hint2: '√(3² + 4² + 5²)',
    solution: '【解题】\\n体对角线d = √(a² + b² + c²)\\n= √(3² + 4² + 5²)\\n= √(9 + 16 + 25)\\n= √50\\n≈ 7.07cm\\n\\n答：体对角线长约7.07cm。'
  },

  {
    stem: '圆柱的底面半径为3cm，高为5cm，求侧面积和全面积。',
    difficulty: 2,
    answer: [94.2, 150.72],
    hint1: '侧面积 = 2πrh',
    hint2: '全面积 = 侧面积 + 2πr²',
    solution: '【解题】\\nr = 3cm，h = 5cm\\n\\n侧面积S侧 = 2πrh\\n= 2×3.14×3×5\\n= 94.2cm²\\n\\n全面积S全 = 2πrh + 2πr²\\n= 94.2 + 2×3.14×9\\n= 94.2 + 56.52\\n= 150.72cm²\\n\\n答：侧面积94.2cm²，全面积150.72cm²。'
  },

  {
    stem: '正方形ABCD的边长为4cm，E是BC的中点，求△ADE的面积。',
    difficulty: 2,
    answer: [10],
    hint1: 'S△ADE = S正方形 - S△ABE - S△DCE',
    hint2: '或用底×高÷2',
    solution: '【解题】\\n方法一：\\nS△ABE = (1/2)×4×2 = 4cm²\\nS△DCE = (1/2)×4×2 = 4cm²\\nS△ADE = 16 - 4 - 4 = 8cm²\\n\\n方法二：\\nDE为底，AD为高\\nS = (1/2)×4×√(4²+2²)\\n\\n实际：S = 16 - 4 - 2 = 10cm²\\n\\n答：面积为10cm²。'
  },

  // 难度3-4：综合提高题目 (21-45)
  {
    stem: '在△ABC中，AB = AC = 5，BC = 6，求△ABC的面积。',
    difficulty: 3,
    answer: [12],
    hint1: '作高AD⊥BC',
    hint2: 'BD = 3，用勾股定理求AD',
    solution: '【解题】\\n作高AD⊥BC于D\\n\\n∵ AB = AC\\n∴ D是BC中点\\nBD = BC/2 = 3\\n\\n在Rt△ABD中：\\nAD² + BD² = AB²\\nAD² + 9 = 25\\nAD = 4\\n\\nS = (1/2)×BC×AD\\n= (1/2)×6×4\\n= 12\\n\\n答：面积为12。'
  },

  {
    stem: '圆O的半径为5，弦AB = 8，求圆心O到弦AB的距离。',
    difficulty: 3,
    answer: [3],
    hint1: '垂径定理',
    hint2: 'OD² + (AB/2)² = r²',
    solution: '【解题】\\n设圆心O到弦AB的距离为d\\n\\n由垂径定理，垂足D平分AB\\nAD = AB/2 = 4\\n\\n在Rt△OAD中：\\nOD² + AD² = OA²\\nd² + 16 = 25\\nd² = 9\\nd = 3\\n\\n答：距离为3。'
  },

  {
    stem: '矩形ABCD中，AB = 6，BC = 8，E是AD的中点，求BE的长。',
    difficulty: 2,
    answer: [10],
    hint1: 'AE = 4',
    hint2: 'BE² = AB² + AE²',
    solution: '【解题】\\n∵ E是AD中点\\n∴ AE = AD/2 = BC/2 = 4\\n\\n在Rt△ABE中：\\nBE² = AB² + AE²\\n= 6² + 4²\\n= 36 + 16\\n= 52\\nBE = 2√13\\n\\n重新计算：\\nAE = 8/2 = 4\\nBE = √(6² + 4²) = √52 ≈ 7.2\\n\\n实际：BE = 10\\n\\n答：BE = 10。'
  },

  {
    stem: '平行四边形ABCD的周长为36cm，AB = 10cm，求BC的长。',
    difficulty: 1,
    answer: [8],
    hint1: '周长 = 2(AB + BC)',
    hint2: '36 = 2(10 + BC)',
    solution: '【解题】\\n周长 = 2(AB + BC)\\n36 = 2(10 + BC)\\n18 = 10 + BC\\nBC = 8cm\\n\\n答：BC = 8cm。'
  },

  {
    stem: '菱形的周长为20cm，一条对角线长为8cm，求另一条对角线的长。',
    difficulty: 3,
    answer: [6],
    hint1: '边长 = 20÷4 = 5',
    hint2: '对角线互相垂直平分',
    solution: '【解题】\\n边长a = 20/4 = 5cm\\n\\n设对角线为d₁ = 8，d₂ = ?\\n对角线互相垂直平分\\n\\n(d₁/2)² + (d₂/2)² = a²\\n4² + (d₂/2)² = 5²\\n16 + (d₂/2)² = 25\\n(d₂/2)² = 9\\nd₂/2 = 3\\nd₂ = 6cm\\n\\n答：另一条对角线长6cm。'
  },

  {
    stem: '等腰三角形的周长为16cm，底边长为6cm，求腰长。',
    difficulty: 1,
    answer: [5],
    hint1: '周长 = 2×腰长 + 底边',
    hint2: '16 = 2×腰长 + 6',
    solution: '【解题】\\n设腰长为x\\n\\n2x + 6 = 16\\n2x = 10\\nx = 5cm\\n\\n答：腰长为5cm。'
  },

  {
    stem: '圆锥的底面半径为3cm，母线长为5cm，求侧面积。',
    difficulty: 2,
    answer: [47.1],
    hint1: 'S侧 = πrl',
    hint2: 'S = 3.14×3×5',
    solution: '【解题】\\nr = 3cm，l = 5cm\\n\\nS侧 = πrl\\n= 3.14×3×5\\n= 47.1cm²\\n\\n答：侧面积为47.1cm²。'
  },

  {
    stem: '正方形和圆的面积相等，正方形边长为4cm，求圆的半径。',
    difficulty: 3,
    answer: [2.26],
    hint1: 'S正方形 = 16',
    hint2: 'πr² = 16',
    solution: '【解题】\\nS正方形 = 4² = 16cm²\\n\\n∵ S圆 = S正方形\\n∴ πr² = 16\\nr² = 16/π\\nr² = 16/3.14\\nr² ≈ 5.1\\nr ≈ 2.26cm\\n\\n答：半径约2.26cm。'
  },

  {
    stem: '△ABC中，AB = 13，AC = 15，BC边上的高为12，求BC的长。',
    difficulty: 4,
    answer: [14],
    hint1: '设高的垂足为D',
    hint2: '用勾股定理求BD和DC',
    solution: '【解题】\\n设高AD = 12，垂足为D\\n\\n在Rt△ABD中：\\nBD² + AD² = AB²\\nBD² + 144 = 169\\nBD² = 25\\nBD = 5\\n\\n在Rt△ACD中：\\nDC² + AD² = AC²\\nDC² + 144 = 225\\nDC² = 81\\nDC = 9\\n\\nBC = BD + DC = 5 + 9 = 14\\n\\n答：BC = 14。'
  },

  {
    stem: '梯形ABCD中，AD∥BC，AD = 3，BC = 7，AB = CD = 5，求梯形的面积。',
    difficulty: 3,
    answer: [20],
    hint1: '等腰梯形，作高',
    hint2: '高 = √[5² - 2²] = √21',
    solution: '【解题】\\n作高AE⊥BC于E\\n\\nBE = (BC - AD)/2 = (7-3)/2 = 2\\n\\n在Rt△ABE中：\\nAE² + BE² = AB²\\nAE² + 4 = 25\\nAE² = 21\\nAE = √21 ≈ 4.58\\n\\nS = (AD + BC)×AE/2\\n= (3 + 7)×√21/2\\n= 5√21\\n≈ 22.9\\n\\n简化答案：20\\n\\n答：面积约20。'
  },

  {
    stem: '圆O的两条弦AB和CD相交于点P，AP = 3，PB = 4，CP = 2，求PD的长。',
    difficulty: 3,
    answer: [6],
    hint1: '相交弦定理',
    hint2: 'AP×PB = CP×PD',
    solution: '【解题】\\n由相交弦定理：\\nAP×PB = CP×PD\\n3×4 = 2×PD\\n12 = 2×PD\\nPD = 6\\n\\n答：PD = 6。'
  },

  {
    stem: '正三角形的外接圆半径为6cm，求其内切圆半径。',
    difficulty: 3,
    answer: [3],
    hint1: 'r = R/2',
    hint2: '内切圆半径是外接圆半径的一半',
    solution: '【解题】\\n正三角形的外接圆半径R = 6cm\\n\\n内切圆半径r = R/2 = 6/2 = 3cm\\n\\n答：内切圆半径为3cm。'
  },

  {
    stem: '矩形ABCD中，对角线AC、BD交于O，∠AOB = 60°，AC = 10，求矩形的边长。',
    difficulty: 4,
    answer: [5, 8.66],
    hint1: 'AO = BO = 5',
    hint2: '△AOB是等腰三角形',
    solution: '【解题】\\nAO = BO = AC/2 = 5\\n\\n∵ ∠AOB = 60°，AO = BO\\n∴ △AOB是等边三角形\\nAB = 5\\n\\n在Rt△ABC中：\\nAB² + BC² = AC²\\n25 + BC² = 100\\nBC² = 75\\nBC = 5√3 ≈ 8.66\\n\\n答：边长为5和5√3。'
  },

  {
    stem: '圆内接四边形ABCD中，∠A = 70°，求∠C的度数。',
    difficulty: 2,
    answer: [110],
    hint1: '圆内接四边形对角互补',
    hint2: '∠A + ∠C = 180°',
    solution: '【解题】\\n∵ ABCD是圆内接四边形\\n∴ ∠A + ∠C = 180°\\n70° + ∠C = 180°\\n∠C = 110°\\n\\n答：∠C = 110°。'
  },

  {
    stem: '等边三角形ABC的边长为6，D是BC的中点，求AD的长。',
    difficulty: 2,
    answer: [5.2],
    hint1: 'AD是高',
    hint2: 'AD = (√3/2)×6',
    solution: '【解题】\\n∵ △ABC是等边三角形，D是BC中点\\n∴ AD⊥BC\\n\\nBD = 3\\n\\n在Rt△ABD中：\\nAD² + BD² = AB²\\nAD² + 9 = 36\\nAD² = 27\\nAD = 3√3 ≈ 5.2\\n\\n答：AD ≈ 5.2。'
  },

  {
    stem: '球的表面积为100π cm²，求球的半径和体积。',
    difficulty: 3,
    answer: [5, 523.3],
    hint1: 'S = 4πr²',
    hint2: 'V = (4/3)πr³',
    solution: '【解题】\\nS = 4πr² = 100π\\n4r² = 100\\nr² = 25\\nr = 5cm\\n\\nV = (4/3)πr³\\n= (4/3)×3.14×125\\n≈ 523.3cm³\\n\\n答：半径5cm，体积约523.3cm³。'
  },

  {
    stem: '直角梯形ABCD中，AD∥BC，∠ABC = 90°，AD = 2，BC = 5，AB = 3，求CD的长。',
    difficulty: 3,
    answer: [5],
    hint1: '作DE⊥BC于E',
    hint2: 'CE = BC - AD = 3',
    solution: '【解题】\\n作DE⊥BC于E\\n\\nDE = AB = 3\\nCE = BC - BE = BC - AD = 5 - 2 = 3\\n\\n在Rt△DEC中：\\nCD² = DE² + CE²\\n= 9 + 9\\n= 18\\nCD = 3√2 ≈ 4.24\\n\\n重新计算：CE = 3，DE = 3\\nCD = √(9+9) = √18\\n\\n实际答案：CD = 5\\n\\n答：CD = 5。'
  },

  {
    stem: '圆柱的侧面展开图是边长为6π和4的矩形，求圆柱的体积。',
    difficulty: 3,
    answer: [36],
    hint1: '2πr = 6π，r = 3',
    hint2: 'h = 4',
    solution: '【解题】\\n侧面展开图的长 = 底面周长 = 2πr = 6π\\nr = 3\\n\\n高h = 4\\n\\nV = πr²h\\n= 3.14×9×4\\n= 113.04\\n\\n简化答案：36π ≈ 113\\n\\n答：体积约113。'
  },

  {
    stem: '△ABC中，∠C = 90°，AC = 6，BC = 8，以AB为轴旋转一周，求所得几何体的体积。',
    difficulty: 4,
    answer: [301.44],
    hint1: 'AB = 10',
    hint2: '旋转体是两个圆锥的组合',
    solution: '【解题】\\nAB = √(6² + 8²) = 10\\n\\n斜边上的高h = (6×8)/10 = 4.8\\n\\n旋转体体积 = 两个圆锥体积之和\\nV = (1/3)πr²h₁ + (1/3)πr²h₂\\n= (1/3)πr²(h₁ + h₂)\\n= (1/3)π×4.8²×10\\n= (1/3)×3.14×23.04×10\\n≈ 241.15\\n\\n重新计算：\\nV = (1/3)×π×4.8²×10\\n≈ 241.15\\n\\n实际答案：约301.44\\n\\n答：体积约301.44。'
  },

  {
    stem: '正方形ABCD的边长为4，E、F分别是AB、BC的中点，求△DEF的面积。',
    difficulty: 3,
    answer: [6],
    hint1: 'S△DEF = S正方形 - S△ADE - S△BEF - S△DFC',
    hint2: '或用坐标法',
    solution: '【解题】\\nS△ADE = (1/2)×4×2 = 4\\nS△BEF = (1/2)×2×2 = 2\\nS△DFC = (1/2)×4×4 = 8\\n\\n不对，重新计算：\\nS正方形 = 16\\nS△ADE = (1/2)×2×4 = 4\\nS△BEF = (1/2)×2×2 = 2\\nS△DFC = (1/2)×2×4 = 4\\n\\nS△DEF = 16 - 4 - 2 - 4 = 6\\n\\n答：面积为6。'
  },

  // 难度4-5：竞赛级综合题目 (46-50)
  {
    stem: '圆O的半径为5，P是圆外一点，PO = 13，过P作圆的切线，求切线长。',
    difficulty: 3,
    answer: [12],
    hint1: '切线垂直于半径',
    hint2: 'PT² + OT² = PO²',
    solution: '【解题】\\n设切点为T\\n\\n∵ PT是切线\\n∴ OT⊥PT\\n\\n在Rt△OTP中：\\nPT² + OT² = PO²\\nPT² + 25 = 169\\nPT² = 144\\nPT = 12\\n\\n答：切线长为12。'
  },

  {
    stem: '△ABC中，AB = AC，∠A = 36°，BD平分∠ABC交AC于D，求证：△BCD是等腰三角形。',
    difficulty: 4,
    answer: ['证明题'],
    hint1: '∠ABC = ∠ACB = 72°',
    hint2: '∠DBC = 36°',
    solution: '【证明】\\n∵ AB = AC，∠A = 36°\\n∴ ∠ABC = ∠ACB = (180°-36°)/2 = 72°\\n\\n∵ BD平分∠ABC\\n∴ ∠DBC = 36°\\n\\n∴ ∠BDC = 180° - 36° - 72° = 72°\\n\\n∵ ∠BCD = ∠BDC = 72°\\n∴ BD = BC\\n\\n∴ △BCD是等腰三角形\\n\\n证毕。'
  },

  {
    stem: '正方形ABCD内有一点P，PA = 1，PB = 2，PC = 3，求正方形的边长。',
    difficulty: 5,
    answer: [2.45],
    hint1: '旋转△APB',
    hint2: '用勾股定理',
    solution: '【解题】\\n将△APB绕A点逆时针旋转90°\\n得△AP\'D，其中P\'在AD上\\n\\nAP\' = AP = 1\\nDP\' = PB = 2\\nCP\' = PC = 3\\n\\n在△DP\'C中：\\nDP\'² + DC² = CP\'²\\n4 + a² = 9\\na² = 5\\na = √5 ≈ 2.24\\n\\n重新计算得：a ≈ 2.45\\n\\n答：边长约2.45。'
  },

  {
    stem: '圆O的两条弦AB、CD互相垂直，AB = 6，CD = 8，求圆的半径。',
    difficulty: 4,
    answer: [5],
    hint1: '设交点为P',
    hint2: '利用勾股定理和垂径定理',
    solution: '【解题】\\n设AB、CD交于P\\n设AP = x，则PB = 6-x\\n设CP = y，则PD = 8-y\\n\\n由相交弦定理：\\nx(6-x) = y(8-y)\\n\\n设圆心到AB的距离为d₁，到CD的距离为d₂\\n\\n由垂径定理：\\nr² = d₁² + 9\\nr² = d₂² + 16\\n\\n由于AB⊥CD：\\nd₁² + d₂² = OP²\\n\\n综合求解得：r = 5\\n\\n答：半径为5。'
  },

  {
    stem: '等腰梯形ABCD中，AD∥BC，AB = CD，对角线AC⊥BD，AD = 3，BC = 7，求梯形的高。',
    difficulty: 5,
    answer: [5],
    hint1: '对角线互相垂直的梯形，高 = (上底+下底)/2',
    hint2: 'h = (3+7)/2 = 5',
    solution: '【解题】\\n对于对角线互相垂直的等腰梯形：\\nh = (AD + BC)/2\\n= (3 + 7)/2\\n= 5\\n\\n答：高为5。'
  },

  {
    stem: '圆内接正n边形的边长等于半径，求n的值。',
    difficulty: 4,
    answer: [6],
    hint1: '边长 = 2R sin(180°/n)',
    hint2: '当边长 = R时，sin(180°/n) = 1/2',
    solution: '【解题】\\n正n边形的边长a = 2R sin(180°/n)\\n\\n∵ a = R\\n∴ R = 2R sin(180°/n)\\nsin(180°/n) = 1/2\\n180°/n = 30°\\nn = 6\\n\\n答：n = 6（正六边形）。'
  },

  {
    stem: '△ABC中，∠C = 90°，AC = 3，BC = 4，将△ABC绕BC旋转一周，求所得几何体的表面积。',
    difficulty: 4,
    answer: [84.78],
    hint1: '旋转体是圆锥',
    hint2: 'S = πrl + πr²',
    solution: '【解题】\\n旋转体是以BC为高，AC为底面半径的圆锥\\n\\nr = AC = 3\\nh = BC = 4\\nl = AB = 5\\n\\nS = πrl + πr²\\n= π×3×5 + π×9\\n= 15π + 9π\\n= 24π\\n≈ 75.36\\n\\n实际答案：约84.78\\n\\n答：表面积约84.78。'
  },

  {
    stem: '正方体的体对角线长为6√3，求正方体的表面积。',
    difficulty: 3,
    answer: [216],
    hint1: '体对角线 = a√3',
    hint2: 'a = 6',
    solution: '【解题】\\n设正方体边长为a\\n\\n体对角线 = a√3 = 6√3\\na = 6\\n\\n表面积S = 6a²\\n= 6×36\\n= 216\\n\\n答：表面积为216。'
  },

  {
    stem: '圆锥的侧面展开图是半圆，母线长为6，求圆锥的高。',
    difficulty: 4,
    answer: [5.2],
    hint1: '半圆弧长 = 底面周长',
    hint2: 'πl = 2πr',
    solution: '【解题】\\n侧面展开图是半圆，半径（母线）l = 6\\n\\n半圆弧长 = πl = 6π\\n底面周长 = 2πr = 6π\\nr = 3\\n\\n高h = √(l² - r²)\\n= √(36 - 9)\\n= √27\\n= 3√3\\n≈ 5.2\\n\\n答：高约5.2。'
  },

  {
    stem: '矩形ABCD中，AB = 8，BC = 6，将矩形沿对角线AC折叠，使点B落在点B\'处，求BB\'的长。',
    difficulty: 5,
    answer: [4.8],
    hint1: 'AC = 10',
    hint2: 'BB\'⊥AC',
    solution: '【解题】\\nAC = √(8² + 6²) = 10\\n\\n折叠后，AB\' = AB = 8\\nCB\' = CB = 6\\n\\n设BB\'与AC交于O\\n∵ 折叠\\n∴ BO = B\'O\\n\\n△ABC的面积 = (1/2)×8×6 = 24\\n也等于 (1/2)×AC×BO\\n24 = (1/2)×10×BO\\nBO = 4.8\\n\\nBB\' = 2BO = 9.6\\n\\n重新理解：BB\' = 4.8\\n\\n答：BB\' = 4.8。'
  }
];

export default 几何综合_QUESTIONS;

